Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, q, tot, head[100001], v[200001], nxt[200001], dep[100001],
    f[100001][17], lg[100001];
void addedge(int x, int y) {
  v[++tot] = y;
  nxt[tot] = head[x];
  head[x] = tot;
}
struct data {
  int rk[11], len;
  data(int len = 0) : len(len) { memset(rk, 0, sizeof(rk)); }
  data operator+(const data &b) const {
    data c(min(len + b.len, 10));
    int l = 1, r = 1;
    for (int i = 1; i <= c.len; i++) {
      if (l <= len && (r > b.len || rk[l] < b.rk[r]))
        c.rk[i] = rk[l++];
      else
        c.rk[i] = b.rk[r++];
    }
    return c;
  }
} rnk[100001][17];
void dfs(int x) {
  for (int i = 1; i <= lg[dep[x]]; i++) {
    f[x][i] = f[f[x][i - 1]][i - 1];
    rnk[x][i] = rnk[x][i - 1] + rnk[f[x][i - 1]][i - 1];
  }
  for (int i = head[x]; ~i; i = nxt[i]) {
    if (v[i] != f[x][0]) {
      dep[v[i]] = dep[x] + 1;
      f[v[i]][0] = x;
      dfs(v[i]);
    }
  }
}
void work(int x, int y, int a) {
  if (dep[x] < dep[y]) swap(x, y);
  data ans;
  int dd = dep[x] - dep[y];
  for (int i = lg[dd]; i >= 0; i--) {
    if ((1 << i) & dd) {
      ans = ans + rnk[x][i];
      x = f[x][i];
    }
  }
  dd = dep[x];
  for (int i = lg[dd]; i >= 0; i--) {
    if (f[x][i] != f[y][i]) {
      ans = ans + rnk[x][i] + rnk[y][i];
      x = f[x][i];
      y = f[y][i];
    }
  }
  ans = ans + rnk[x][0];
  if (x != y) ans = ans + rnk[f[x][0]][0] + rnk[y][0];
  printf("%d ", min(ans.len, a));
  for (int i = 1; i <= min(ans.len, a); i++) printf("%d ", ans.rk[i]);
  printf("\n");
}
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 2; i <= n; i++) lg[i] = lg[i >> 1] + 1;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    addedge(x, y);
    addedge(y, x);
  }
  for (int i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    if (rnk[x][0].len < 10) rnk[x][0].rk[++rnk[x][0].len] = i;
  }
  dep[1] = 1;
  dfs(1);
  for (int i = 1; i <= q; i++) {
    int x, y, a;
    scanf("%d%d%d", &x, &y, &a);
    work(x, y, a);
  }
}